This Coste-Shiota result contradicts what I thought I knew about Thom's career.  Isn't the point that there are examples where you just can't improve "topological stability" to "smooth stability"?  (see Todd's comment for a simple explanation.)

Here is a famous example of Whitney's.  Take in R^3 the four hypersurfaces x = 0, y = 0, x = y, and x = zy.  Let's restrict our attention to where 0 < z < 1.  Stratify so that the union of these hypersurfaces is the 2-skeleton and the line x = y = 0 is the 1-skeleton.  Consider the map $(x,y,z) \mapsto z$.  (Maybe we'd prefer a proper map, so require $x^2 + y^2 \leq 1$ as well and stratify the boundary in the obvious way.)  The fibers of this map are R^2's (or disks) stratified by four lines through the origin.  One of those lines moves around as z changes.

This map is a submersion on each stratum, so by Thom's isotopy lemma the fibers are all homeomorphic in a stratum-preserving way.  But any such trivialization must fail to be C^1 along z = 0, since any C^1 map from one fiber to another has to preserve the cross ratio of those four lines.